Answer:
10
Step-by-step explanation:
Ground level is where h = 0, so solve the equation ...
h(x) = 0
-5(x -4)^2 +180 = 0 . . . . substitute for h(x)
(x -4)^2 = 36 . . . . . . . . . . divide by -5, add 36
x -4 = 6 . . . . . . . . . . . . . . positive square root*
x = 10 . . . . . . add 4
The object will hit the ground 10 seconds after launch.
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* The negative square root also gives an answer that satisfies the equation, but is not in the practical domain. That answer would be x = -2. The equation is only useful for time at and after the launch time: x ≥ 0.
<span>I rounded up the whole number as if each student sells only 3 sweatshirts, it is too little to pay for the whole trip, therefore the number must be 4. Hope that helps!</span>
It looks like the given equation is
sin(2x) - sin(2x) cos(2x) = sin(4x)
Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
which lets us rewrite the equation as
sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)
Move everything over to one side and factorize:
sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0
sin(2x) - 3 sin(2x) cos(2x) = 0
sin(2x) (1 - 3 cos(2x)) = 0
Then we have two families of solutions,
sin(2x) = 0 or 1 - 3 cos(2x) = 0
sin(2x) = 0 or cos(2x) = 1/3
[2x = arcsin(0) + 2nπ or 2x = π - arcsin(0) + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
(where n is any integer)
[2x = 2nπ or 2x = π + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
[x = nπ or x = π/2 + nπ]
… … … or [x = 1/2 arccos(1/3) + nπ or x = -1/2 arccos(1/3) + nπ]
Answer:
You did not provide options but it would be any one of them that starts with -3 and then has a decimal.